Let's Make Robots!

How to select a Motor?

Hey guys!

I'm Thinking to build an all terrain bot in the next few weeks. I've all ready built a few bots which were not that great. Every time something gets messed up, like the drive system or the wheels. So, I Dont want to make the same mistakes again and decided to get help from u people. How to select a motor generally? I mean is there any calculation to know the motor you select will be powerful enough to achieve certain conditions? By conditions i refer to the velocity of the bot, max inclination it can traverse etc.

And finally Which type of drive system is best suited for locomotion in  All terrain bots?

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Sorry! See Below

Most motors spin very fast and have very little torque. For most practical robot building purposes, you are going to want to slow the motor down with gearing. Many motors come with an attached gearbox (often referred to as 'gearmotors' or 'gearhead motors'), or you can use external gearing.

You can find many gearmotors on robot hobby shops online stores. Often the description will provide you with some guidance on its use (e.g., for a small fast robot).

Your motor/gear train combo need enough torque to do whatever job you need. Torque is measured in a length x length unit (in-ounces, N-cm, etc). This is the amount of force that the motor/gear train provide at a set radial distance from the center of the axle. In other words, if your motor puts out 90 in-ounces of torque, and you are using it to move a robot with 1 inch wheels, the motor can handle 90 ounces.

The rated RPM is also important, since that will relate to how fast your motor spins. Note that RPM is rated 'unloaded', and will be slower once the motor is doing some work.

Both RPM and torque are rated at a certain voltage.

I Still have a few doubts. Lets consider the motor you mentioned in your example, a motor rated 90 in-ounces of torque say at 100 rpm. What do mean by the motor can 'handle' 90 ounces when using 1 inch wheel? Does that mean the motor can move a bot that weighs 90 ounces at the rated rpm, in this case 100 rpm?

What happens if I increase the diameter of the wheel? The velocity of the bot in m/s can be calulated using this formula (pi* d* n)/60 rite?

What ignoble means is that with a 90 in.oz motor, and a 1" radius wheel, the force produced at the edge of the wheel will be equivalent to 90 ounces.
If the bot in question was winching itself vertically upwards on a rope wrapped around this wheel, then the bot would have to weigh less than 90 ounces (ignoring friction from air/bearing/etc, and assuming no slippage). If the bot is instead driving around on the ground, it would be able to tackle any incline (no-slip condition again) without overstressing the motor provided it weighted less than 90 ounces. As the weight of the bot increases beyond 90 ounces you start to get a decreasing limit of the maximum incline it can drive up - for example if the bot weighed 127 ounces it would be limited to slopes of 45° or less, which is still probably steeper than the grip from the wheels would allow anyway.
Also don't forget to add up your torques if you've got multiple motors driving together!

Your formula is correct - the forward velocity of the bot will increase proportionally as you increase the wheel diameter, but using the same motor. However it's important to remember that the force produced by the wheel will decrease proportionally as you increase the wheel diameter, so the weight/incline the motor can handle will drop.

Thanks Telefox, now I Understand his reply. Is there any relation to find out how much inclination a bot can travel?

What if the drive system is Belt drive/ chain drive instead of wheels? Should the diameter of the sprockets/pulleys be considered instead of the diameter of the wheel?

The portion of a vehicle's weight that the drive system has to overcome at the wheel-ground interface is (vehicle weight)*(sine(angle)). Once again this ignores any kind of frictional losses - you'll need to allow for, let's say, a minimum of 30% extra torque.

Depending on how you connect the transmission together there may be some gearing relationships that need to be accounted for. If you examine the path from the motor to the ground does it go:
• Motor shaft -> sprocket -> chain -> sprocket -> ground (i.e. the second sprocket is in contact with the ground).
• Motor shaft -> sprocket -> chain -> sprocket -> wheel shaft -> wheel -> ground (i.e. the second sprocket is connected to a wheel that has a larger diameter).

When a small gear drives a large gear, or a small sprocket is chained to a larger sprocket, the gearing relationship reduces the output rotational speed but increases the output torque, with the change directly proportional to the difference in effective diameters. The opposite is also true.
If two wheel/gears/sprockets are locked onto the same shaft, they'll have the same rotational speed and torque values.

With those few points you should be able to work out virtually any transmission's basic properties.

Got the Point about sprockets telefox. Thanks a ton! I will be connecting the sprockets directly to the motor shafts.

Sorry for my Ignorrance, But I dont understand the formula used to calculate the max inclination. Say the bot weighs around 100 ounces. I have a motor which puts out 130 ou-in of torque. If i use two of those motors I get a equivalent torque of about 260 ou-in. So what will be the max inclination the bot can travel?


Ok, first we'll add 30% to the weight of the bot, which should hopefully cover the losses in a modest system, so the bot's effective weight is now 130 ounces.
The force the motors have to overcome, due to the bot's weight, is going to be a maximum of 130 ounces. Since the motors together produce 260 in.oz of torque, they'll produce (260 in.oz)/r worth of force, where 'r' is the wheel's radius in inches. As long as the radius is less than 2" you'll get 130 ounces of forward thrust, which is enough to overcome any incline.

If the wheel's radius is now greater than 2", you'll start to see a decreasing limit on the max inclination. To calculate backwards for the max angle, you use the rearranged formula: angle = sin-¹((260/r)/130).

As an example, using the above formula for r=3": angle = sin-¹((260/3)/130)
                                                                       angle = sin-¹(86.666/130)
                                                                       angle = 41.81° (maximum)

Thanks again telefox. Now I get the relation. Helped a lot :)